Answer:
Option B
Explanation:
$\left(216\right)^{\frac{3}{5}}\times \left(2500\right)^{\frac{2}{5}}\times \left(300\right)^{\frac{1}{5}}=$ $\left(3^{3}\times 2^{3}\right)^{\frac{3}{5}}\times \left(5^{4}\times 2^{2}\right)^{\frac{2}{5}}\times \left(5^{2} \times 2^{2} \times 3\right)^{\frac{1}{5}}$ $=3^{\left[3\times\frac{3}{5}\right]}\times2^{\left[3\times\frac{3}{5}\right]}\times5^{\left[4\times\frac{2}{5}\right]}\times2^{\left[2\times\frac{2}{5}\right]}\times5^{\left[2\times\frac{1}{5}\right]}\times2^{\left[2\times\frac{1}{5}\right]}\times 3^{\frac{1}{5}}$ $=3^{\frac{9}{5}} \times 2^{\frac{9}{5}} \times 5^{\frac{8}{5}} \times 2^{\frac{4}{5}} \times 5^{\frac{2}{5}} \times 2^{\frac{2}{5}} \times 3^{\frac{1}{5}}$ $=3^{\left[\frac{9}{5}+\frac{1}{5}\right]}\times 2^{\left[\frac{9}{5}+\frac{4}{5}+\frac{2}{5}\right]}\times 5^{\left[\frac{8}{5}+\frac{2}{5}\right]}$ $=3^{2}\times 2^{3}\times 5^{2}$ Hence, the number of prime factors = $\left(2+3+2\right)=7$
